0
$\begingroup$

I need to express the following in term of the unit step function:

$ f(t) = \begin{cases} 0& 0\le t<3\\ -2& 3\le t<5\\ 2& 5\le t<7\\ 1& t\ge 7\end{cases} $

My solution:

$ f_1(t) = 0 $

$ f_2(t) = 0 - 2u_2(t) $

$f_3(t) = 0 - 2u_2(t) +4u_3(t) $

$f_4(t) = 0 - 2u_2(t) +4u_3(t) - 1u_4(t) $

$ f(t) = - 2u_2(t) +4u_3(t) - u_4(t) $

wondering if my working is correct

$\endgroup$
  • 1
    $\begingroup$ Why is the first jump at $2$? Why is the second jump at $3$? Why is the last jump at $4$? It seems to me that the jumps should be at $3$, $5$, and $7$. $\endgroup$ – Ian Jul 28 '16 at 20:50
  • $\begingroup$ why is that???? $\endgroup$ – mp12345 Jul 28 '16 at 20:54
1
$\begingroup$

Your coefficients are correct but as stated by Ian in the comments, you are using the wrong Heaviside functions, you should be jumping at the intervals 3,5 and 7 because that is where your piecewise function changes value:

\begin{equation} f(t)=-2u_3(t)+4u_5(t)-u_7(t) \end{equation}

Check:

\begin{equation} f(8)=-2+4-1=1 \\ f(6)=-2+4=2 \end{equation}

etc.

$\endgroup$
  • $\begingroup$ see link for the definition of the heaviside function $\endgroup$ – Decebalus Jul 28 '16 at 21:04
1
$\begingroup$

If $u(t)$ is the unit step function, $f(t)$ can be written as

$$f(t)=-2u(t-3)+4u(t-5)-u(t-7)$$

$\endgroup$
  • $\begingroup$ any explanation for this ansswer? thanks for your time $\endgroup$ – mp12345 Jul 28 '16 at 21:02
  • 1
    $\begingroup$ @javahelp: Well, $u(t-3)$ implements the jump at $t=3$, $u(t-5)$ jumps at $t=5$, and $u(t-7)$ jumps at $t=7$. The factors take care of the height of the jump. First we go to $-2$ then up by $+4$ (resulting in $+2$), then again down by $1$, leaving us with a constant value of $1$ for $t>7$. $\endgroup$ – Matt L. Jul 28 '16 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.